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x^2+42x=392
We move all terms to the left:
x^2+42x-(392)=0
a = 1; b = 42; c = -392;
Δ = b2-4ac
Δ = 422-4·1·(-392)
Δ = 3332
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3332}=\sqrt{196*17}=\sqrt{196}*\sqrt{17}=14\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-14\sqrt{17}}{2*1}=\frac{-42-14\sqrt{17}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+14\sqrt{17}}{2*1}=\frac{-42+14\sqrt{17}}{2} $
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